Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Helium atom is two times heavier than a hydrogen molecule. At 289 K, the average kinetic energy of a helium atom is

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two times that of a HYDROGEN molecule
same as that of a hydrogen molecule
four times that of a hydrogen molecule
half that of a hydrogen molecule.

Solution :Average kinetic ENERGY DEPENDS only UPON TEMPERATURE
Discussion
2.

Helium and hydrogen show exceptional behavour in deviation from ideal gas nature. Why ?

Answer»

Solution :Hydrogen and HELIUM are lighest GASES. Molecules of these gases have ALMOST negligible attractive FORCES among them. Therefore they behave exceptionally.
Discussion
3.

Heisenberg principle is applicable to microscopic particles.

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ANSWER :T
Discussion
4.

Heavy water is used as a ………… in nuclear reactors.

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ANSWER :MODERATOR
Discussion
5.

Heavy water is used as a

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FUEL in engines
semiconductor
moderator in NUCLEAR reactors
insulator in STEAM engines.

Answer :C
Discussion
6.

Heavy water is used aas

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DRINKING water
detergent
washing water
a moderator

Answer :D
Discussion
7.

Heavy water is used as

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MODULATOR in nuclear REACTIONS
 coolant in nuclear reactions
both (a) and (B)
none of these

Solution :HEAVY water is used as moderator as well as coolant in nuclear reactions.
Discussion
8.

Heavy water is toxic to micro-organism because of

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Its HIGH MOLECULAR WEIGHT
Lesser chemical reactivity
Its hygroscopicnature
GREATER density

Solution :Absorbs water
Discussion
9.

Heavy water is too micro-organisms because of

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Its HIGH molecular weight
Lesser CHEMICAL reactivity
Its hygroscopic nature
Greater density

Solution :ABSORBS WATER
Discussion
10.

Heavy water is present in

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Rain water
Remains of HIMALAYAN SNOW melt
LEAVES of Banyan TREES
All the above

Answer :D
Discussion
11.

Heavy water is obtained by....

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boiling water.
fractional DISTILLATION of `H_2O`
PROLONGED electrolysis of `H_2O`
heating `H_2O_2`

Answer :C
Discussion
12.

Heavy water is obtained by

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BOILING water
heating `H_(2)O_(2)`
PROLONGED electrolysis of `H_(2)O`
all of these.

Solution :`D_(2)O` is prepared by repeated electrolysis of ordinary water CONTAINS a small amount of alkali.
Discussion
13.

What happen when granulated zinc is reacted with NaOH ?

Answer»
Discussion
14.

Heavy water is found in minute quantities in rain water, on the leaves of Banyan trees, and in the last remain obtain by melting of snow as in himalayas. The number of neutrons in deuterium is

Answer»

2
3
1
0

Solution :`D = ""_1H^(2), N = 2-1 = 1`
Discussion
15.

Describe Clark's method to remove hardness of water ?

Answer»
Discussion
16.

Heavy water is left as a residue by carrying out the repeated electrolysis of ordinary water. Infact, its boiling point (374.42K) is slightly more than that of ordinary water (373K). It is very expensive and this can be judged by the fact about 29000 litres water on repeated electrolysis leave behindone litre of heavy water (D_(2)O). (i) How is heavy water different from ordinary water is its biological action? (ii) Give the chemical reactions of heavy water with Al_(4)C_(3) and CaC_(2)). (iii) What are the values associated with the use of heavy water?

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SOLUTION :(i) Heavy water is injouries to human being, animals and plants since it slows down the rate of reaction that OCCURS in them. Ordinary water is no way injurious.
(ii) `Al_(4)C_(3)+12D_(2)P to 4Al(OD)_(3)+4Al(OD)_(3)+UNDERSET("Deuteromehtane")(CD_(4))`
`CaC_(2)+2D_(2)O toCa(OD)_(2)DC-=+underset("Deuteromehtane")(CD_(4))`
Discussion
17.

Heavy water is mainly used as a

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MEDICINE
DETERGENT
COOLENT
MODERATOR

ANSWER :D
Discussion
18.

Heavy water is found in minute quantities in rain water, on the leaves of Banyan trees, and in the last remain obtain by melting of snow as in himalayas. The weight precentage fo deuterium in heavy water is

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0.02
0.5
0.2
0.1

Solution :MOLECULAR weight of `D_2O = 20`
4 grams of D in 20 gramsof `D_2O`
THEREFORE percentage of `D= 4/20 xx 100 = 20`
Discussion
19.

Heavy water is found in minute quantities in rain water, on the leaves of Banyan trees, and in the last remain obtain by melting of snow as in himalayas. Which of the following represents the heavy water ?

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WATER at 227K
water containinglarge CONTAMINATION of LEAD salts
deuterium oxide
protium oxide

SOLUTION :`D_2 O`
Discussion
20.

Heavy water is

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`""^(1)H_(2)O^(16)`
`""^(2)H_(2)O^(16)`
`""^(3)H_(2)O^(16)`
`""^(3)H_(2)O^(18)`

Answer :B
Discussion
21.

Heavy water is …..

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WATER CONTAINING FE, Cr, Mn
Water at `0^@C`
`D_2O`
`H_2O^18`

SOLUTION :`D_2O` in which `D=._1H^2`
Discussion
22.

Heavy water is _________.

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WATER containing `FE, CR,Mn`
water at 0`.^(@)C`
`D_(2)O`
`H_(2).^(18)O`

ANSWER :C
Discussion
23.

Heavy water is ____________

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De-mineralilzed WATER
De-ionized water
ordinary water containing DISSOLVED salts of heavy metals.
The compound of heavier ISOTOPE of hydrogen with oxygen.

Answer :D
Discussion
24.

Heavy water (D_(2)O) is

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A product of oxygen and hydrogen
WATER of MINERAL springs
Water obtained by repeated distillation and condensation
Ordinary water containing DISSOLVED SALTS HEAVY metals.

Answer :C
Discussion
25.

Heavy water freezes at

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`0^@C`
`3.8^@C`
`-0.38^@C`
`-4^@C`

ANSWER :B
Discussion
26.

What is super heavy water ?

Answer»
Discussion
27.

Heavy water can be prepared by

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FRACTIONAL DISTILLATION
EXHAUSTIVE electrolysis
By EXCHANGE reactions
All the above

Answer :D
Discussion
28.

Heats of ionisation of acetic acid and ammonium hydroxide are x kJ mol^(-1) and y kJ mol^(-1). Heat of neutralisation of HCl and NaOH is z kJ mol^(-1). Calculate the heat when acetic acid is neutralised with ammonium hydroxide. (z-(x+y)kJ mol^(-1))

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Solution :`(Z- (X +y) kJ MOL ^(-1))`
Discussion
29.

Heats of formation ofCH_(3)COOH(l), CO_(2)(g) and H_(2)O(l) are respectively -487, -394 and -286 kJ mol^(-1). Calculate the heat of combustion of acetic acid.

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ANSWER :`-873KJmol^(-1) `
Discussion
30.

Heats of combustion of graphite carbon and carbon monoxide are respectively - 393.5 kJ mol^(-1) and - 283 kJ mol^(-1). Calculate the heat of formation of carbon monoxide.

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Solution :`C_(("GRAPHITE")) + O_(2(g)) to CO_(2(g)) , DELTAH = -393.5 kJ "" …(1)`
`CO_((g)) + 1/2O_(2(g)) to CO_(2(g)) , DeltaH = -283kJ ""....(2)`
Equation (1) - Equation (2), gives
`C_(("graphite")) + 1/2O_(2(g)) to CO_((g)), DeltaH = -110.5 kJ`.
Discussion
31.

The enthalpies of combustion of carbon and carbon monoxide are -390 KJ mol^(-1) and -278 KJ mol^(-1) respectively. The enthalpy of formation of carbon monoxide is

Answer»

Solution :`C _("graphicle") + O _(2 (G)) to CO _(2 (g)) , DELTA H =- 393.5 kJ……..(1)`
`CO _((g)) + (1)/(2) O _(2 (g)) to CO _(2 (g)) , Delta H =- 283 Kj ……..(2)`
Equation (1) - equation (2) , gives
`C _(("graphicle")) + (1)/(2) O _(2 (g)) to CO _((g)) ‘ Delta H =-110 .5 kJ`
Discussion
32.

Heats of combustion of garphite carbon and carbon monoxide are respectively -393.5 KJ mol^(-1) and -110.5 "mol"^(-1). Then determine the bond enthalpies.

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Solution :`C_(("graphite"))+O_(2)(g) to Co_(2)(g):DeltaH=-393.5KJ....(1)`
`CO_((g))+ 1/2O_(2)(g) to CO_(2)(g) , DeltaH=-110.5KJ....(2)`
EQUATION (1) - equation (2) ,gives
`c_(("graphite"))+1/2O_(2)(g) to Co_((g)),DeltaH=-283KJ`
The HEAT of FORMATION of carbon monoixde =-392.5-(-110.5)=-283KJ `mol^(-1)`.
Thermochemical equation,`c_(("graphite"))+1/2O_(2)(g) to Co_((g)),DeltaH=-283KJ`.
Discussion
33.

Heats of combustion of C_2H_4, H_2 and C_2H_6 are -1409 K.J, -285 K.J and -1558 KJ. Heat of hydrogenation of ethylene is

Answer»

`-136 KJ`
`-240 KJ `
`136 KJ`
` 227 KJ`

ANSWER :A
Discussion
34.

Heats of combustion diamond carbon and garphite carbon are respectively -395.4 KJ mol^(-1) and -393.5 KJ mol^(-1). Calculate the heat of transition of dimand carbon into garphite carbon.

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Solution :`C_(("diamond"))+O_(2)(G) to O_(2)(g) ,DeltaH=-395.4KJ....(1)`
`C_(("GRAPHITE"))+ O_(2)(g) to O_(2)(g) ,DeltaH=-395.4KJ....(2)`
EQUATION (1)-euation (2), GIVES `C_(("diamond"))toC_(("graphite")),DeltaH=-1.9KJ`
The heat TRANSITION of diamond carbon into graphite carbone =-1.9KJ.
Discussion
35.

Heats of atomisation of NH_(3) and NH_(2)H_(4) are xKJmol^(-1)" and "yKJmol^(-1)respectively. What is the average energy ofN-N bond ?

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ANSWER :`Y-1.33xKJmol^(-1)`
Discussion
36.

Heating of 2-chloro-l-phenylbutane with EtOK // EtOH gives X as the major product. Reaction of X with Hg(OAc)_(2) // H_(2)O followed by NaBH_(4) gives Y as the major product. Y is :

Answer»




SOLUTION :
Discussion
37.

Heating mixture of Cu_(2)O and Cu_(2)S will give

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CU + `SO_2`
Cu + `SO_3`
CUO + CUS
`Cu_(2)SO_3`

SOLUTION :N//A
Discussion
38.

Heating an ore in the absence of air below its melting point is called

Answer»

Leaching
Roasting
Smelting
CALCINATION

SOLUTION :Heating an ORE in the ABSENCE of AIR below its melting point is called calcination.
Discussion
39.

Heating an aqueous solution of aluminium chloride to dryness will give

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`AlCl_(3)`
`Al_(2)Cl_(6)`
`Al_(2)Cl_(3)`
`Al(OH)Cl_(3)`

ANSWER :B
Discussion
40.

Heating a mixture of sodium benzoate and soda-lime gives

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Benzene
Methane
Sodium BENZOATE
CALCIUM benzoate.

ANSWER :A
Discussion
41.

Heating an ammoniacal solution of MgSO_(4) in the presence of NH_(4)Cl and Na_(2)HPO_(4) causes the precipitation of:

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`Mg (HPO_(4))`
`Mg (NH_(4))PO_(4)`
`MgCl_(2)`
`Mg (NH_(4))_(2) (PO_(4))_(2)`

SOLUTION :A white ppt of `Mg(NH_(4)) PO_(4)` is formed
Discussion
42.

Heating an aqueous solution of aluminium chloride to dryness will give :

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`AICI_3`
`Al_(2)CI_6`
`Al_(2)O_(3)`
`Al(OH)Cl_(2)`

ANSWER :C
Discussion
43.

Heat of vaporization of benzene is 7350 cal K^(-1) "mol"^(-1). Calculate the change in entropy for converting 1 mole gaseous benzene to liquid benzene at 77^(@) C.

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21 CALORIE `K^(-1) "MOL"^(-1)`
`-21 "calorie" K^(-1) "mol"^(-1)`
`-21 "calorie" K^(-1)`
`21 "calorie" K^(-1)`

Answer :B
Discussion
44.

Heat released in neutralisation of 1 mole of HF with excess NaoH is

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57.32kJ
`GT 57.32kJ`
`LT 57.32kJ`
none of these

Solution :For HF, `Delta H_("neutralization") gt 57.3`
Discussion
45.

Heat of reaction (DeltaH) is given by

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`Delta H =` ACTIVATION energy of forward reaction - Activation energy of backward reaction
`Delta H=` Sum of BOND energy of reactants Sum of bond energy of PRODUCTS
`Delta H=` Sum of ENTHALPY of products - Sum of enthalpy of reactants
All the above

Answer :D
Discussion
46.

Heat of neutralization of a strong acid by a strong base is a constant value because ____

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SOLUTION :only `H^+` and `OH^-` IONS react in every case
Discussion
47.

Heat of neutralization is always _____

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SOLUTION :NEGATIVE
Discussion
48.

Statement -I: Heat of neutralization of HCl and NaOH is same as that of H_(2)SO_(4) with NaOH. Statement-II : HCl, H_(2)SO_(4) and NaOH are all strong electrolyte.

Answer»

`57.32KJ `
`GT 57.32KJ`
`LT 57.32KJ`
NONE of these 

ANSWER :B
Discussion
49.

Heat of neutralisation of H_2SO_4 with NaOH is - x KJ. Then the heat of ionisation of water is

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`-X gt KJ`
`x KJ`
`gt x KJ`
`LT x KJ`

ANSWER :B
Discussion
50.

Heat of neutralisation of a weak acid HA by NaOH is -"12.13 kJ mol"^(-1). Calculate the enthalpy of ionization of HA. The Heat of neutralisation of a strong acid with strong base is -"54.9 kJ mol"^(-1)

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Solution :`DeltaH_("IONISATION")^(SIGMA)=+"42.77 KJ mol"^(-1)`.
Discussion